Two charges are placed at a certain distance apart in the air. If a glass slab is introduced between them, then the force between the two charges will

1.  Become zero

2.  Decrease

3.  Increase

4.  Remain the same

Subtopic:  Coulomb's Law |
 71%
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A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is

1.  qEy

2.  qEy2

3.  qE2y

4.  q2Ey

Subtopic:  Electric Field |
 83%
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Force of interaction between two bodies of the same nature of the charge 

1.  Must be repulsive

2.  May be attractive

3.  Must be attractive

4.  Zero

Subtopic:  Coulomb's Law |
From NCERT
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On the axis of a uniformly charged ring of radius R, as shown, the electric field is maximum at

1.  x = 0

2.  x = R2

3.  x = R2

4.  x = R2

Subtopic:  Electric Field |
 74%
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At the three corners of an equilateral triangle, charges are placed as shown in the figure. The magnitude of the electric field at point O is

1.  14πε02Qr2

2.  14πε0Qr2

3.  14πε0Q2r2

4.  14πε0Q3r2

Subtopic:  Electric Field |
 57%
From NCERT
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Two symmetrical parallel metal each of one side plate area A having charges Q1 and -Q2 are placed at small separation. What will be the electric field at a point in between the plates?

1.  Q1+Q22Aε0

2.  Q1+Q24Aε0

3.  Q1-Q22Aε0

4.  Q1-Q24Aε0

Subtopic:  Electric Field |
 56%
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An electric dipole of the moment P is released in a uniform electric field E from the position of maximum torque. The angular speed of the dipole when P becomes parallel to E will be [l = moment of inertia of dipole]

1.  2PEI

2.  PEI

3.  4PEI

4.  2IPE

Subtopic:  Electric Dipole |
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A charge Q is given to a conducting sphere of radius R. Now if a charge -Q is placed, as shown, at a distance r from the center, then the magnitude of the force of attraction between charges is

1.  14πε0Q2r2

2.  <14πε0Q2r2

3.  >14πε0Q2r2

4.  14πε0Q2r-R2

 

Subtopic:  Coulomb's Law |
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If the electric flux entering and leaving a closed surface are respectively of magnitude ϕ1 and ϕ2, then the electric charge inside the surface will be

1.  ε0ϕ2-ϕ1

2.  ϕ1-ϕ2/2ε0

3.  ε0ϕ2+ϕ1

4.  ϕ2-ϕ1ε0

Subtopic:  Gauss's Law |
 62%
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Electric field at an axial point of short dipole is  E1. If the electric field at the equatorial point of same dipole is E2, then which of the following is correct?

1.  E1 . E2 < 0

2.  E1 . E2 > 0

3.  E1 . E2 = 0

4.  E1 . E2 = 0

Subtopic:  Electric Dipole |
 51%
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